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共2个回答
热心网友
夹逼定理
lim [n^2/(n+n^2)]<原极限<lim [n^2/(1+n^2)]
且lim [n^2/(n+n^2)]=lim [n^2/(1+n^2)]=1
所以
原极限=1
热心网友
这个需要使用夹逼准则来求解:
因为: 1/(n + n^2) ≤ 1/(m + n^2) ≤1/(1 + n^2), 1 ≤ m ≤ n
所以: n * 1/(n + n^2) = n/(n + n^2) ≤∑1/(m + n^2) ≤ n * 1/(1 + n^2) = n/(1 + n^2)
由于:lim n * n/(n + n^2) = lim n^2/(n + n^2) = lim 1/(1 + 1/n) = 1
lim n * n/(1 + n^2) = lim n^2/(1 + n^2) = lim 1/(1 + 1/n^2) = 1
因此,lim n*∑(1/(m + n^2) = 1